In an adiabatic process, the system is insulated from the surroundings and heat absorbed or released is zero. From Eq. (12.1), we see that work done by the gas results in decrease in its internal energy (and hence its temperature for an ideal gas).

We quote without proof (the result that you will learn in higher courses) that for an adiabatic process of an ideal gas.

`color{purple} { P V^ γ = "const"}` ......................... (12.13)

where γ is the ratio of specific heats (ordinary or molar) at constant pressure and at constant volume.

where `γ` is the ratio of specific heats (ordinary or molar) at constant pressure and at constant volume.

`color{green} {gamma = (C_p)/C_v}`

Thus if an ideal gas undergoes a change in its state adiabatically from `(P_1, V_1)` to `(P_2, V_2)` :

`color{purple} {P_1 V_1^γ = P_2 V_2^ γ}` ........................ (12.14)

Figure12.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two isotherms.



We can calculate, as before, the work done in an adiabatic change of an ideal gas from the state `color{green} {(P_1, V_1, T_1)}` to the state `color{purple} { (P_2, V_2, T_2)}`.

`color{navy} { W = int_(V_1) ^(V_2) P dV`

`= "constant" xx int_(V_1) ^(V_2) (dV)/V^(gamma) = "constant" xx (V^(- gamma + 1))/(1 - gamma)|_(V_1)^(V_2)`

`= ("constant")/(1 -gamma) xx { 1/(V_2^(gamma -1)) - 1/ (V_1^(gamma -1)) ]` ....................12.15

From Eq. (12.34), the constant is `P_1V_1^γ` or `P_2V_2^γ`

`W = 1/(1 -gamma) [ ( P_2V_2^gamma)/(V_2^(gamma -1) - P_1V_1^gamma)/ ( V_1^(gamma -1 ) ) ]`

`color{purple} {= 1/ (1- gamma ) [ ( P_2V_2 - P_1V_1 ] = (mu R(T_1 - T_2 ))/(gamma -1) ` ......................12.16

As expected, if work is done by the gas in an adiabatic process (W > 0), from Eq. (12.16), `T_2 < T_1`. On the other hand, if work is done on the gas (W < 0), we get `T_2 > T_1` i.e., the temperature of the gas rises.

In an isochoric process, V is constant.

No work is done on or by the gas. From Eq. `ΔQ = ΔU + ΔW`............(12.1),
the heat absorbed by the gas goes entirely to change its internal energy and its temperature.

The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant volume.

In an isobaric process, P is fixed. Work done by the gas is

`color{orange} {W = P (V_2 – V_1) = μ R (T_2 – T_1)}` ............... (12.17)

Since temperature changes, so does internal energy. The heat absorbed goes partly to increase internal energy and partly to do work.

The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant pressure.

In a cyclic process, the system returns to its initial state. Since internal energy is a state variable, `ΔU = 0` for a cyclic process.

From Eq. (12.1), the total heat absorbed equals the work done by the system.